Answer:
HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)
NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)
Explanation:
A buffer is a solution that resists changes in acidity or alkalinity. A buffer is able to neutralize a little amount of acid or base thereby maintaining the pH of the system at a steady value.
A buffer may be an aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid.
The equations for the neutralizations that occurred upon addition of HCl or NaOH are;
HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)
NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)
Answer:

Explanation:
Hello!
In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

It means we can compute the final density as shown below:

Now, we plug in to obtain:

Regards!
Answer:
V = 240.79 L
Explanation:
Given data:
Volume of butane = ?
Temperature = 293°C
Pressure = 10.934 Kpa
Mass of butane = 33.25 g
Solution:
Number of moles of butane:
Number of moles = mass/ molar mass
Number of moles = 33.25 g/ 58.12 g/mol
Number of mole s= 0.57 mol
Now we will convert the temperature and pressure units.
293 +273 = 566 K
Pressure = 10.934/101 = 0.11 atm
Volume of butane:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
V = nRT/P
V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm
V = 26.49 L/0.11
V = 240.79 L
Answer:
6,1,2
Explanation:
_Ag + _N2 => _Ag3N
LHS RHS
Ag=1×6=3 Ag=3×2=6(Balanced)
N=2×1=2 N=1×2=2(Balanced)
coefficient=6,1,2
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